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anonymous
 one year ago
can't understand why integral of xe^((x^2)/2) is e^((x^2)/2)
can someone explain how to calculate this integral please?
anonymous
 one year ago
can't understand why integral of xe^((x^2)/2) is e^((x^2)/2) can someone explain how to calculate this integral please?

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[\LARGE xe^{\frac{x^2}{2}}\] this one?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3do you know usubstitution?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3The usubstitution is an integration method that essentially involves using the chain rule in reverse.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3hmm let's try rewriting that fraction \[\LARGE xe^{\frac{1}{2}x^2}\] so we can see the exponent clearly.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3we need to pick a u so that when we take that derivative of u we can substitute.. so what will be our u? Well we do see an x in front of the e.. Maybe we can have \[\LARGE u=\frac{1}{2}x^2 \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so what is the derivative of \[\large \frac{1}{2}x^2 ? \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3we can use product rule to make this a bit easier. Let f(x) = 1/2 and g(x) = x^2 recall that the product rule is f(x)g'(x) +f'(x)g(x)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so what is the derivative of f(x) = 1/2 ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3similarly what is the derivative of \[\large g(x) = x^2 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so if \[u= (1/2)x^2 > u' = x\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3yeah but we have to write this differently it will be du = x dx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so now we have \[xe^u dx\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3hmm how do I explain this.... we just have \[e^u \] and substitute the u back .. don't forget to add +C

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3remember du = x dx ? since there's nothing on the left hand side.... oh man .. I'm having trouble explaining this part. . . usually when there's nothing ... it's just blank like 1 sorry I'm in panic mode I can't figure out how to type this well

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3but we just have \[\LARGE e^u +C \] now substitute \[\LARGE u=\frac{1}{2}x^2 \] back into \[\LARGE e^u +C \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so \[xe^u dx > xe^u (1/x) dy > e^u du\] (+c...)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3you don't have to write dx you don't have to write du

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3du = x dx it's just .... @ganeshie8 can you help me explain this? I know what's going on just can't put it into words

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, I think I understand, thank you very much!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3but in the end you should have \[\LARGE e^u +C \] now plug in u back into this equation

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3and once we have our antiderivative, we can simply check by taking the derivative.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3you should have something like \[\LARGE e^{\frac{1}{2}x^2} +C\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3now take the derivative of that equation...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3that's how to check :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we get back to \[e^((x^2)/2) \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3if you take the derivative of your antiderivative, we should be getting that x in front of that e

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I wrote the integral

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[\LARGE f(x) = e^{\frac{1}{2}x^2} +C \] \[\LARGE f'(x) = xe^{\frac{1}{2}x^2} +C \] because the derivative is \[\frac{1}{2}x^2 \rightarrow \frac{1}{2} (2x) + 0(x^2) \rightarrow x\] and that's placed in front of the e.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3the derivative of any e^x is always in the front and the exponent part of e^x just stays there :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Excellent. Thanks a lot!
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